There are many ways to check if a sequence of estimators is consistent.

By definition, a sequence of estimators $W_n = W_n(X_1,X_2,\ldots,X_n)$ is a consistent sequence of estimators of the parameter $\theta$ if for every $\epsilon > 0$ $$\lim_{n\to\infty}\mathbf P_\theta (|W_n-\theta| < \epsilon) = 1$$

While it is not necessary that $\lim_{n\to\infty} E_\theta[(W_n-\theta)^2] = 0$, we can be sure that $W_n$ is a consistent sequence of estimators if $\lim_{n\to\infty} E_\theta[(W_n-\theta)^2] = 0$. This is due to Chebyshev's inequality: $$0 \le \mathbf P_\theta (|W_n-\theta| \ge \epsilon) \le \frac{E_\theta[(W_n-\theta)^2]}{\epsilon^2}$$ If $E_\theta[(W_n-\theta)^2] \to 0$ as $n\to\infty$, $\mathbf P_\theta (|W_n-\theta| \ge \epsilon) \to 0$ as $n\to \infty$, and we obtain $\lim_{n\to\infty}\mathbf P_\theta (|W_n-\theta| < \epsilon) = 1$.

Question: Since (apparently) $\lim_{n\to\infty} E_\theta[(W_n-\theta)^2] = 0$ is only a sufficient condition for $W_n$ to be a consistent sequence of estimators, and not a necessary one, how do I produce a consistent sequence of estimators $W_n$ for which $\lim_{n\to\infty} E_\theta[(W_n-\theta)^2]\ne 0$? If we cannot do this, the condition must be necessary and sufficient, which is stronger. Would appreciate any help, thank you!


1 Answer 1


Maximum likelihood estimators in logistic regression are one example. The MLE is infinite with non-zero probability, which stays non-zero for all $n$, though it decreases exponentially with $n$. $E[(W_n-\theta)^2]$ is infinite, and does not converge to zero.

If you don't like estimators that can be non-finite, take $W_n$ as mean of a sample of size $n$ from a $t_2$ distribution. The distribution has a well-defined mean, zero, but no finite variance, so $E[(W_n-\theta)^2]=E[W_n^2]$ is always infinite.

I don't know of any naturally occurring examples where $E[(W_n-\theta)^2]$ is finite but doesn't converge to zero, but it's easy to rig up examples if $W_n$ is allowed to be silly. For example, if you have $n$ iid observations from $N(\mu,1)$ and take $W_n=\bar X_n$ with probability $1-1/n$ and $W_n=n^{2/3}$ with probability $1/n$, then $W_n$ is consistent for $\mu$ (because $P(W_n=\bar X_n)\to 1$) but the variance doesn't vanish.

Basically, what you need is that a vanishing fraction of 'wrong' values of $W_n$ can be very very wrong, but most of the values of $W_n$ are close to $\theta$.


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